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I am working on the following problem from Gerald Teschl's book on ODE's and am at a loss of how to proceed.

Suppose $U=\mathbb{R} \times \mathbb{R}^n$ and that $|f(t,x)| \leq g(|x|)$ for some positive continuous function $g \in C([0,\infty))$ which satisfies $$\int_0^{\infty} \frac{dr}{g(r)} = \infty$$ Then all solutions of the IVP $f(t,x) = \dot{x}$, $x(0) = x_0$ are defined for all $t \geq 0$.
Show that the same conclusion still holds if there is such a function $g_T(r)$ for every $t \in [0,T]$.
(Hint: Look at the differential equation for $r(t)^2 = |x(t)|^2$.)

I am not sure how to use the hint and I have not been successful in any of my attempts at the problem. Any help would be appreciated. Thanks!

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1 Answer 1

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Assume that the maximal interval of definition of the solution is $[0,T)$. We have to prove that $T=+\infty$.

Let $r(t) = |x(t)|$ and assume for the moment that $x(t) \neq 0$ for every $t\in [0,T)$, so that $r$ is of class $C^1$ and $$ \dot{r}(t) = \frac{\dot{x}(t) \cdot x(t)}{|x(t)|} \leq |\dot{x}(t)| = |f(t, x(t))| \leq g(r(t)). $$ Since $g$ is a positive function you have that $$ \frac{\dot{r}(s)}{g(r(s))} \leq 1, \qquad \forall s\in [0,T). $$ Integrating this inequality on $[0,t]$ you get $$ \int_0^t \frac{\dot{r}(s)}{g(r(s))}\, ds \leq t, \qquad \forall t\in [0,T). $$ The first integral can be computed with the change of variable $y = r(s)$, obtaining $$ (1)\qquad \int_{r_0}^{r(t)} \frac{1}{g(s)}\, ds \leq t, \qquad \forall t\in [0,T). $$ Assume now, by contradiction, that $T < +\infty$. In this case you must have $r(t) = |x(t)| \to +\infty$ for $t \to T^-$. But this is in contradiction with (1), since the l.h.s. diverges by assumption to $+\infty$ whereas the r.h.s. goes to $T$.

The assumption $x(t) \neq 0$ for every $t\in [0,T)$ is not restrictive. Namely, since we are assuming by contradiction that $|x(t)| \to +\infty$ for $t\to T^-$, then we have that there exists some $t_0\in [0,T)$ such that $|x(t)| > 0$ for every $t\in [t_0, T)$. Now it is enough to reason as above on the interval $[t_0, T)$.

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  • $\begingroup$ Why does the l.h.s have to diverge? What if $g(s) = s^2$? $\endgroup$ Sep 19, 2017 at 13:28
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    $\begingroup$ The assumption is $\int_0^\infty 1/g(s) \, ds = +\infty$, with $g\in C([0,\infty))$, $g > 0$. $\endgroup$
    – Rigel
    Sep 19, 2017 at 19:52

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